以下哪一种技术是整数除2的最佳选择,为什么?
技巧1:
x = x >> 1;
技巧2:
x = x / 2;
这里x是整数。
以下哪一种技术是整数除2的最佳选择,为什么?
技巧1:
x = x >> 1;
技巧2:
x = x / 2;
这里x是整数。
当前回答
使用除法(/),假设这样更清楚。编译器将相应地进行优化。
其他回答
我说这些是为了参加编程比赛。一般来说,他们有非常大的输入,除以2会发生很多次,已知输入是正的或负的。
X >>1比X /2好。我在ideone.com上运行了一个程序,其中发生了超过10^10除以2的运算。X /2花了将近5.5s,而X >>1花了将近2.6s。
Mod 2, test for = 1。不知道c中的语法,但这可能是最快的。
我想说有几件事需要考虑。
Bitshift should be faster, as no special computation is really needed to shift the bits, however as pointed out, there are potential issues with negative numbers. If you are ensured to have positive numbers, and are looking for speed then I would recommend bitshift. The division operator is very easy for humans to read. So if you are looking for code readability, you could use this. Note that the field of compiler optimization has come a long way, so making code easy to read and understand is good practice. Depending on the underlying hardware, operations may have different speeds. Amdal's law is to make the common case fast. So you may have hardware that can perform different operations faster than others. For example, multiplying by 0.5 may be faster than dividing by 2. (Granted you may need to take the floor of the multiplication if you wish to enforce integer division).
如果您追求的是纯粹的性能,我建议您创建一些可以执行数百万次操作的测试。对执行进行多次采样(您的样本量),以确定哪一个在统计上最适合您的操作系统/硬件/编译器/代码。
X/Y是正确的…和" >> "移位运算符..如果我们想要二除一个整数,我们可以使用(/)被除数运算符。移位运算符用于移位位。
x = x / 2; x / = 2;我们可以这样用…
就CPU而言,位移运算比除法运算快。 然而,编译器知道这一点,并将适当地优化到它可以的程度, 因此,您可以以最有意义的方式编写代码,并且知道您的代码是有意义的 有效地运行。但是请记住,由于前面指出的原因,在某些情况下,无符号整型可以比整型更好地优化。 如果你不需要符号算术,那么不要包含符号位。